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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.1, 10 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={ β–ˆ(π‘₯+1, 𝑖𝑓 π‘₯β‰₯1@&π‘₯2+1 , 𝑖𝑓 π‘₯<1)─ Since we need to find continuity at of the function We check continuity for different values of x When x < 1 When x = 1 When x > 1 Case 1 : When x < 1 For x < 1, f(x) = π‘₯^2+1 Since this a polynomial It is continuous ∴ f(x) is continuous for x < 1 Case 2 : When x = 1 f(x) is continuous at π‘₯ =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)= 𝑓(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) ((1βˆ’β„Ž)^2+1) = (1βˆ’0)^2+1 = 1^2+1 = 1 + 1 = 2 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) (1+β„Ž)+1 = lim┬(hβ†’0) 2+β„Ž = 2 + 0 = 2 & 𝑓(1) = π‘₯+1 = 1+1 = 2 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x=1 Case 3 : When x > 1 For x > 1, f(x) = x + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence, there is no point of discontinuity Thus, f is continuous for all π’™βˆˆπ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.